Съдържание

BPO 2024, Теория — Задача 1

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Всички материали от BPO 2024

BPO6 Problems 20-24 June 2024. Montenegro · 20 юни 2024 г. · 25 т.

Условие

A small tennis ball resting on top of a large basketball; the bottom of the basketball is at height h above the ground (orange line), marked with a double arrow labelled h.

A vertical stack of four balls of decreasing size labelled B1 (bottom, largest), B2, B3, B4 (top), with the label n=4; the bottom of B1 is at height h above the ground (orange line), marked with a double arrow labelled h.

A) A tennis ball with the mass $m_2$ sits on top of a basketball with the mass $m_1$ . The bottom of the basketball is a height $h$ above the ground, and the bottom of the tennis ball is at a height $h + d$ above the ground (see the first figure). The balls are dropped to the ground. To what height does the tennis ball bounce?

Note: Work in the approximation where $m_1$ is much larger than $m_2$ and assume that the balls bounce elastically.

B) Now consider $n$ balls, $B_1, \cdots, B_n$ , having masses $m_1, m_2, \cdots, m_n$ (with $m_1 \gg m_2 \gg \cdots \gg m_n$ ), sitting in a vertical stack (see the second figure). The bottom of $B_1$ is at a height $h$ above the ground, and the bottom of $B_n$ is a height $h + l$ above the ground. The balls are dropped. In terms of $n$ , to what height does the top ball bounce? If $h = 1.0\ \mathrm{m}$ , what is the minimum number of balls needed for the top one to bounce to a height of at least $1.0\ \mathrm{km}$ ? The same question, if the top ball is to reach the Earth’s escape speed?

Note: Work in the approximation where $m_1$ is much larger than $m_2$ , which is much larger than $m_3$ , etc., and assume that the balls bounce elastically. Ignore wind resistance, etc., and assume that $l$ is negligible.

Data: $l$ is not very large, $R_{Earth} = 6.4 \cdot 10^{6}\ \mathrm{m}$ , $g = 9.8\ \mathrm{m/s^2}$ , .

Отговори

Решение

Lets examine collision of B1 and B2

$-m_1 \upsilon + m_2 \upsilon = m_1 \upsilon_1 + m_2 \upsilon_2$ (yy΄) (1)

$\tfrac{1}{2}\, m_1 \upsilon^2 + \tfrac{1}{2}\, m_2 \upsilon^2 = \tfrac{1}{2}\, m_1 \upsilon_1^2 + \tfrac{1}{2}\, m_2 \upsilon_2^2$ (2)

[2 points]

Solving the system

$\upsilon_1 = [(3m_2 - m_1)/(m_2 + m_1)]\, \upsilon \quad \upsilon1 = [(1 - 3\, m_1/m_2)/(1 + m_1/m_2)]\, \upsilon$

but $m_1/m_2 = 0$ and $\upsilon_1 = 3\upsilon$

$I/2\, m_1 (3\upsilon)^2 = m_1 g H$

$\tfrac{1}{2}\;9 \upsilon^2 = g H$

$H = \tfrac{1}{2}\;9 \upsilon^2/g = 9h$

And the answer d+9h

[3 points]

Lets examine collision of B2 and B3

$-m_3 \upsilon + m_2 3\upsilon = m_3 \upsilon_3 + m_2 \upsilon_2$ (yy΄) (1)

$\tfrac{1}{2}\, m_3 \upsilon^2 + \tfrac{1}{2}\, m_2 9\upsilon^2 = \tfrac{1}{2}\, m_3 \upsilon_3^2 + \tfrac{1}{2}\, m_2 \upsilon_2^2$ (2)

Solving the system

$\upsilon_3 = 7\upsilon$

[2 points]

In general

$\upsilon_n = (2^n - 1)\upsilon$

[8 points]

$H = l + ((2^n - 1)\upsilon)^2/2g = l + (2^n - 1)^2 h$

[1 point]

If h is 1 meter, and we want this height to equal 1000 meters, then (assuming l is not very large) we need $2^n - 1 > \surd\, 1000$ . Five balls won’t quite do the trick, but six will, and in this case the height is almost four kilometers.

[4 points]

Escape velocity from the earth (which is $\upsilon_{esc} = \surd\, 2gR \approx 11,200\ \mathrm{m/s}$ ) is reached when

$\upsilon_n \geq \upsilon_{esc} \Longrightarrow (2n - 1)\ \mathrm{sqrt}\,(2gh) \geq \mathrm{sqrt}\,(2gR) \Longrightarrow n \geq \log_2(\mathrm{sqrt}\,(R/h) + 1)$

With $R = 6.4 \bullet 10^{6}\ \mathrm{m}$ and $h = 1\ \mathrm{m}$ , we find $n > 12$ .

[5 points]

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf